Concepts Of Physics MCQ Edition [Volume 1]PhysicsCentre of Mass, Linear momentum, Collision
A block of mass m is positioned on a triangular block of mass M , which in turn rests on a horizontal surface as shown in the figure. Assuming all surfaces are frictionless, determine the velocity of the triangular block when the smaller block reaches the bottom end.
Options
- A[ 2M^2 gh ^2 (M + m)(m + M ^2 ) ]^ 1/2
- B[ 2m^2 gh ^2 (M + m)(M + m ^2 ) ]^ 1/2
- C[ 2m^2 gh ^2 (M + m)(M - m ^2 ) ]^ 1/2
- D[ 2m^2 gh ^2 (M + m)(M + m ^2 ) ]^ 1/2
Correct answer
D. [ 2m^2 gh ^2 (M + m)(M + m ^2 ) ]^ 1/2
Step-by-step solution
Let V be the velocity of the triangular block of mass M towards the left. Let v_r be the velocity of the block of mass m relative to the triangular block, directed down the incline. The velocity of the block m relative to the ground is v = v _r + V . Taking the right direction as positive x and upward as positive y , the velocity vectors are: V = -V i v _r = v_r i - v_r j v = (v_r - V) i - v_r j Since there is no external force in the horizontal direction, the horizontal momentum of the system is conserved. Initial